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Electric Potential and Voltage

The electric potential is the potential energy per unit charge. When we put into an electric field a charge of value qq, the electric potential is equal to:

φ=Epq.\varphi = \frac{E_p}{q}.

The unit of electric potential is volt (VV).

The work the electric field does is equal to:

W=∫CFe⋅ds=−Wext,W = \int_C \boldsymbol{F_e} \cdot d\boldsymbol{s} = -W_{ext},

where WextW_{ext} is the external work. The work needed to move a particle.

The potential energy is equal to the work required to move a charge qq from an infinite distance to a position described by the displacement vector r\boldsymbol{r}:

Ep=Wext=−∫CFe⋅ds=−∫CqE⋅ds. \begin{align*} E_p = W_{ext} &= -\int_C \boldsymbol{F_e} \cdot d\boldsymbol{s} \\ &= -\int_C q \boldsymbol{E} \cdot d\boldsymbol{s}. \\ \end{align*}

The separation vector s\boldsymbol{s} described by radial coordinates is equal to:

s(t)=trxrx^+tryry^,s(t)=t,s^(t)=rxrx^+ryry^,s′(t)=rxrx^+ryry^,r≤t≤∞. \begin{align*} \boldsymbol{s}(t) &= t \frac{r_x}{r} \boldsymbol{\hat{x}} + t \frac{r_y}{r} \boldsymbol{\hat{y}}, \\ s(t) &= t, \\ \boldsymbol{\hat{s}}(t) &= \frac{r_x}{r} \boldsymbol{\hat{x}} + \frac{r_y}{r} \boldsymbol{\hat{y}}, \\ \boldsymbol{s'}(t) &= \frac{r_x}{r} \boldsymbol{\hat{x}} + \frac{r_y}{r} \boldsymbol{\hat{y}}, \\ r &\leq t \leq \infty. \end{align*}

Then the electric potential is equal to:

φ=−1q∫CqE⋅ds,=−∫CE⋅ds,=−∫∞rkeQs(t)2s^⋅s′(t)dt=keQ∫∞r−1t2(rxrx^+ryry^)⋅(rxrx^+ryry^)dt=keQ∫∞r−1t2dt=keQ[1t]∞r=keQr. \begin{align*} \varphi &= -\frac{1}{q} \int_C q \boldsymbol{E} \cdot d\boldsymbol{s}, \\ &= - \int_C \boldsymbol{E} \cdot d\boldsymbol{s}, \\ &= -\int_{\infty}^r k_e \frac{Q}{s(t)^2} \boldsymbol{\hat{s}} \cdot \boldsymbol{s'}(t) dt \\ &= k_e Q \int_{\infty}^r -\frac{1}{t^2} \left(\frac{r_x}{r} \boldsymbol{\hat{x}} + \frac{r_y}{r} \boldsymbol{\hat{y}}\right) \cdot \left(\frac{r_x}{r} \boldsymbol{\hat{x}} + \frac{r_y}{r} \boldsymbol{\hat{y}}\right) dt \\ &= k_e Q \int_{\infty}^r -\frac{1}{t^2} dt \\ &= k_e Q \left[\frac{1}{t}\right]_{\infty}^r \\ &= \frac{k_e Q}{r}. \end{align*}

By the gradient theorem:

E=−∇φ,\boldsymbol{E} = - \boldsymbol{\nabla} \varphi,

meaning the electric field points into the direction of steepest decrease in electric potential.

The voltage is the difference in potentials:

U=φ2−φ1=keQ(1r2−1r1). \begin{align*} U &= \varphi_2 - \varphi_1 \\ &= k_e Q \left(\frac{1}{r_2} - \frac{1}{r_1}\right). \end{align*}

It is also the line integral of electric field:

U=keQ(∫∞r2−1t2dt−∫∞r1−1t2dt)=keQ(∫∞r2−1t2dt+∫r1∞−1t2dt)=keQ∫r1r2−1t2dt=−∫CE⋅ds, \begin{align*} U &= k_e Q \left(\int_{\infty}^{r_2} -\frac{1}{t^2} dt - \int_{\infty}^{r_1} -\frac{1}{t^2} dt\right) \\ &= k_e Q \left(\int_{\infty}^{r_2} -\frac{1}{t^2} dt + \int_{r_1}^{\infty} -\frac{1}{t^2} dt\right) \\ &= k_e Q \int_{r_1}^{r_2} -\frac{1}{t^2} dt \\ &= - \int_C \boldsymbol{E} \cdot d\boldsymbol{s}, \\ \end{align*}

or work per unit charge:

U=−1q∫CFe⋅ds=−Wq=Wextq, \begin{align*} U &= -\frac{1}{q} \int_C \boldsymbol{F_e} \cdot d\boldsymbol{s} \\ &= -\frac{W}{q} \\ &= \frac{W_{ext}}{q}, \end{align*}

where CC is the path from r1r_1 to r2r_2 and WextW_{ext} is the work required to move the particle along path CC.

A point of charge QQ is creating a radial electric field. Calculate the electric potential at point AA.

Radial potential illustration

The displacement between AA and QQ is equal to:

r=A−Q,\boldsymbol{r} = A - Q,

then:

φ=keQr.\varphi = \frac{k_e Q}{r}.

What is the work required to move a charge qq from a point with potential φ1\varphi_1 to a point with potential φ2\varphi_2?

The work can be derived from the voltage:

U=Wextq=φ2−φ1,Wext=q(φ2−φ1). \begin{align*} U &= \frac{W_{ext}}{q} = \varphi_2 - \varphi_1, \\ W_{ext} &= q (\varphi_2 - \varphi_1). \end{align*}
Wext=1.5⋅10−2 J.W_{ext} = 1.5 \cdot 10^{-2}\ J.